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If one knows the remainder of dividing a number $n$ by $2$ and $3$, is it possible to find and argument to know the remainder of dividing $n$ by $6$ without having to resort to the Chinese remainder theorem?

The converse (finding the remainders of dividing $n$ by $2$ and $3$ knowing the other one) is easy, but I'm not seeing a way to do this without using CRT.

Jyrki Lahtonen
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Mark_Hoffman
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    Why do you want to avoid the very useful (and easy to understand) theorem ? – Peter Nov 05 '22 at 08:59
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    You can of course detect the bijection by taking the expressions $6k,6k+1,6k+2,6k+3,6k+4,6k+5$ modulo $2$ and $3$ , but this is basically the CRT for the case $(2,3)$ – Peter Nov 05 '22 at 09:02
  • A proof'll be a special case of a CRT proof, e.g. by Easy CRT in the dupe, if $,\color{#c00}{m\equiv 1\pmod{m'}},$ then

    $$\begin{align} n&\equiv a!!\pmod{m'}\ n&\equiv b!!\pmod{m}\end{align}\iff n \equiv,\underbrace{ b+m\left[\dfrac{a-b}{\color{#c00}m}\color{#c00}{\bmod m'}\right]}_{\textstyle \bbox[5px,border:1px solid #c00]{ ma + (1-m):!b}}\pmod{mm'}\qquad$$ OP is special case $,m',m = 2,3.\ \ $

     – Bill Dubuque Nov 05 '22 at 10:57
  • The point is - as explained here - the Easy CRT formula is very easy to compute since one modulus $,\color{#c00}{m},$ is $\color{#c00}{\equiv 1}$ mod the other $,\color{#c00}{m'},$ so the fraction (inverse) $,(a-b)/m\equiv a-b \pmod {m'}$ in the Easy CRT formula is trivial to compute, so Easy CRT is even easier in this case. – Bill Dubuque Nov 05 '22 at 11:08
  • If you know basic ring theory then it is possible to view this formula more conceptually in terms of the orthogonal idempotents yielding the direct product (Peirce) decomposition, e.g. see here and here. – Bill Dubuque Nov 05 '22 at 11:17

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